32% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five, inclusive

There are two possible answer for the question asked, yes(more likely to make purchases) or no. The probability for a random adults answer yes to the question is 32% so the probability that the adults answer no is 68%.

(a) exactly two

There is only one case for this question, 2 people say yes and 8 people say no. The calculation for this problem will be:

2P10 * P(yes)^2 * P(no)^8= 10!/2!8!* 32%^2 * 68%^8 = 0.21066= 21.06%

(b) more than two

There are a lot of case for more than two, its easier to find out the negation of the probability which is "two or less". Case that fulfill "two or less" will be: 2 yes and 8 no

1 yes and 9 no

0 yes and 10 no

The calculation for the negation will be:

~P(X)= 0P10 * P(yes)^0 * P(no)^10 + 1P10 * P(yes)^1 * P(no)^9 + 2P10 * P(yes)^2 * P(no)^8 =

10!/0!10!* 32%^0 * 68%^10 + 10!/1!9!* 32%^1 * 68%^9 + 10!/2!8!* 32%^2 * 68%^8

0.02113 + 0.09947 + 0.21066 = 0.33126= 33.12%

Since its the negation, you need to subtract 1 with the negation

P(X)= 1 - ~P(X)

P(X)= 1 - 33.12%= 66.88%

(c) between two and five, inclusive

Same formula as above, but the case is:

2 yes and 8 no

3 yes and 7 no

4 yes and 6 no

5 yes and 5 no

The calculation will be:

2P10 * P(yes)^2 * P(no)^8 + 3P10 * P(yes)^3 * P(no)^7 + 4P10 * P(yes)^4 * P(no)^6 + 5P10 * P(yes)^5 * P(no)^5 =

10!/2!8!* 32%^2 * 68%^8 + 10!/3!7!* 32%^3 * 68%^7 + 10!/4!6!* 32%^4 * 68%^6 + 10!/5!5!* 32%^5 * 68%^5

0.21066 + 0.26435 + 0.21770 + 0.1229= 0.81561= 81.56%